Solve. displaystyleintfrac{{{78}{x}}}{{{25}-{12}{x}+{4}{x}^{2}}}

Question
Integrals
asked 2020-12-15
Solve. \(\displaystyle\int\frac{{{78}{x}}}{{{25}-{12}{x}+{4}{x}^{2}}}\)

Answers (1)

2020-12-16
The given integration is
\(\displaystyle\int\frac{{{78}{x}}}{{{25}-{12}{x}+{4}{x}^{2}}}\)
A simple calculation the integration becomes
\(\displaystyle\int\frac{{{78}{x}}}{{{25}-{12}{x}+{4}{x}^{2}}}={7}\int\frac{{{\left.{d}{x}\right.}}}{{{25}-{12}{x}+{4}{x}^{2}}}\)
\(\displaystyle={7}\int\frac{{{\left.{d}{x}\right.}}}{{{4}{x}^{2}-{12}{x}+{25}}}\)
\(\displaystyle={7}\int\frac{{{\left.{d}{x}\right.}}}{{{4}{\left({x}^{2}-{3}{x}+\frac{25}{{4}}\right)}}}\)
\(\displaystyle=\frac{7}{{4}}\int\frac{{{\left.{d}{x}\right.}}}{{{x}^{2}-{3}{x}+\frac{25}{{4}}}}\)
\(\displaystyle=\frac{7}{{4}}\int\frac{{{\left.{d}{x}\right.}}}{{{x}^{2}-{3}{x}+\frac{25}{{4}}}}\)
\(\displaystyle=\frac{7}{{4}}\int\frac{{{\left.{d}{x}\right.}}}{{{x}^{2}-{2}\cdot{x}\cdot\frac{3}{{2}}+\frac{9}{{4}}+\frac{25}{{4}}-\frac{9}{{4}}}}\)
\(\displaystyle=\frac{7}{{4}}\int\frac{{{\left.{d}{x}\right.}}}{{{\left({x}-\frac{3}{{2}}\right)}^{2}+\frac{16}{{4}}}}\)
\(\displaystyle=\frac{7}{{4}}\int\frac{{{\left.{d}{x}\right.}}}{{{\left({x}-\frac{3}{{2}}\right)}^{2}+{\left({2}\right)}^{2}}}\)
\(\displaystyle=\frac{7}{{4}}\times\frac{1}{{2}}{{\tan}^{ -{{1}}}{\left(\frac{{{x}-\frac{3}{{2}}}}{{2}}\right)}}+{C}\)
\(\displaystyle{\left[\text{By using the formula}\ \int\frac{{{\left.{d}{x}\right.}}}{{x}^{2}}+{a}^{2}=\frac{1}{{a}}{{\tan}^{ -{{1}}}{\left(\frac{x}{{a}}\right)}}+{C}\right]}\)
\(\displaystyle=\frac{7}{{8}}{{\tan}^{ -{{1}}}{\left(\frac{{{2}{x}-{3}}}{{4}}\right)}}+{C}\)
Where C is the constant of integration.
Thus, the value of the integration is
\(\displaystyle\int\frac{{{7}{\left.{d}{x}\right.}}}{{{25}-{12}{x}+{4}{x}^{2}}}=\frac{7}{{8}}{{\tan}^{ -{{1}}}{\left(\frac{{{2}{x}-{3}}}{{4}}\right)}}+{C}\)
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